BAM!!! Mr. Tarrou. So we are finally going
to learn where we do all this math and apply it to the real world. Now you will be doing
this kind of math, at least in this PreCalculus class or certainly Physics and Engineering,
where all this trig that we are learning actually does have a real life application. Even if
it hidden by the fact that computers do so much of this work for us now. So this is only
introduction to… an introduction to vectors. This is only PreCalculus class that I am teaching.
But you know… Hopefully these examples I am about to go over, there is going to be
5 examples of applications of vectors. Putting some real world settings and looking at some
forces that are acting within that setting. Not sure if it is going to be sophisticated
enough for some of your physics lessons if that is what you are watching this video for.
But I hope that they do help. So we have an example here, A wagon is pulled by a handle
making a 40 degree angle with a force of 35 pounds. What amount of force is actually moving
the wagon along the level ground? So let’s give ourselves a quick little picture. Not
very pretty here, but a couple of wheels… a wagon… and a handle. Ok, and up that handle
is being applied a force of 35 pounds. Well, that handle… that force is being applied..
It is being pulled. You are walking and this wagon is following you along. That force is
being applied along an angle of… What do we have here? 40 degrees. So, not all of that
35 pounds is being used to pull the wagon. Actually with this angle you are lightning
the front wheels a little bit. You are wasting some energy in the fact that… you know…
at this angle…. We are only wasting energy in the sense that we are trying to move it
along level ground. So what is that actual force? What is the force moving this wagon
along the level ground? Well, I am putting an x here but when we talk about vectors my
textbook really stresses the horizontal and vertical components. Our vectors are going
to be written ‘a’ times the component vector i plus ‘b’ times component vector j. I will
switch back and forth between this component notation and this shorthand notation of just
simplyThat is why I loose or use so much chalk. I keep breaking it into pieces.
We are going to find that horizontal component of this vector. This vector here acting as
the handle. We have this vector going at 40 degrees with 35 pounds. How do you find the
horizontal component? Well this is just simply going to be a small right triangle. And so
thus, how do you find the sides of a right triangle. You just simply do your SOHCAHTOA.
Based on this angle what do you want and what do you have? I want the adjacent side and
I have the hypotenuse. So the cosine of 40 is equal to the horizontal component of a
over the magnitude of the vector which is 35 pounds. And we get ‘a’ is equal to 35 times
the cosine of 40. And 35cos(40) is equal to 26.8 pounds. So if you are pulling that handle
with 35 pounds of force, actually only 26.8 pounds of that are helping to move it horizontally
along a horizontal plane. Now it is not part of this question, but if you want to know
how much of your force is lifting up and is waisted force you have… If you wanted to
find the vertical component of ‘b’, then… well… let’s see. Based on this 40 degree
angle, b is the opposite side, and the 35 pounds is the hypotenuse… so the sine of
40 is equal to b over 35 pounds. And ‘b’ is equal to 35 times the sine of 40 degrees.
That ends up being 22.25 pounds of force. So you again exerting 35 pounds of force to
pull that wagon we have some force in a horizontal component and some force in a vertical component.
Moving that wagon was 26.8 pounds. Now what if you wanted to move this wagon a distance
of 400 feet? How much work would be done? Well, there are a couple of versions of this
formula. But what I am going to work with is Work is calculated by doing the amount
of force that is applied in this direction, or actually the amount of force being applied
times the cosine of the angle times the magnitude or distance of movement. So the force is being
applied is 35 pounds. The cosine of the angle… so the cosine of 40 degrees… times the amplitude.
Now let’s take a second here and look at what the force times the cosine represents. The
force being applied is 35 pounds again, but that is not how much energy is being used
to move the wagon horizontally along the flat ground or horizontal plane. So it kind of
makes sense that we are doing 35 times the cosine of 40 because I just worked through
that same idea when I was trying to find the horizontal force being applied on that wagon.
So that was just basically the horizontal component we just found a minute ago. So 35
times the cosine of 40, or the magnitude of force times the cosine of theta times the
magnitude of ab… how far you want to move it. We want to move that wagon 400 feet. That
answer comes out to… taking out my notes here. I did not work that out. Well let’s
go find that answer shall we! Alright, sorry about that. We have a total of 10,724.6 foot/pounds
of work. Alright! Let’s go onto another example. Alright! The magnitude and direction of two
forces on an object are 50 pounds at a bearing of south 27 degrees west and 80 pounds at
a bearing of north 70 degrees west. What is the magnitude and direction of the resultant
vector? Now I am kind of giving us a little bit of an advanced problem right off the bat.
Like I didn’t just give you two vectors and asked you to find the resultant. Because if
I just give you two vectors and asked you to find the resultant, if they are already
in component form you just add the a’s and add the b’s and then you get your answer.
This one is a bit difficult because of the fact that we are given bearings and in my
current PreCalculus book it does not do a lot with bearings. It just sprinkle them in
enough my students never seem to be completely comfortable with them. And also there are
no drawings. So what we are going to do is… we are going to get this drawn. We are going
to get our vectors from magnitude and direction into their x and y components… their horizontal
and vertical components. Then add them together, get a resultant vector, and then convert that
back into magnitude and direction so we can actually answer the question. What is the
final magnitude and the direction of the resultant vector? Well drawing these word problems is
always a big struggle. So let’s get that drawn and see what this looks like. So we have a
North South line. And we have a vectors. They are starting from the same point. They are
going to start right there. South 27 degrees West. Well, here is South and West is to the
left. We are going to do that by 27 degree. So let’s take a look at maybe drawing something
like this. Ok, now that is going to be vector v. It is South 27 degrees West. So there is
a little angle in here of 27 degrees. That is South 27 degrees West. We have another
vector which we are going to highlight in blue. It is 80 pounds at a bearing of North
70 degrees West. So North and 70 degrees West. I am going to make this a little bit longer
because there is more force behind it. I am going to run out of chalkboard as well! We
have an 80 pound force. It is at North 70 degrees West. And the bearing of the other
one by the way, we will call this w. The magnitude of vector v is 50. Ok. So we have this drawn.
Let’s get a representation of what the resultant is going to look like. We are going to be
looking for, let me finish my parallelogram here that we are starting this drawing off
with…. We are going to find the resultant vector that is down here. If I have drawn
this somewhat reasonable it looks like there is a downward trend to the left. So this resultant
vector which is going to be w+v, or v+w, is going to be in quadrant 3. We will need to
keep that in mind when we find a final direction of this vector. Ok, well I can’t do anything
until I have my vectors in their component form… the horizontal and vertical components.
And this is the formula which I showed in a earlier video of how it works, where it
comes from, that we are going to use to do this. So vector v is going to be its magnitude
which is 50 times the cosine of the… uh… Cosine of the angle. See our calculators don’t
understand bearings. You know… It understands standard position rotation. So I need not
just the force of each vector, the 80 and 50 pounds, but I also need to know the standard
position rotation. Remember standard position rotation is off of the x axis. So we have
got a rotation of what? For vector w that rotation starts from here, goes 90 and then
an additional 70 degrees. So the angle for w is going to be equal to 90 plus 70 or 160
degrees. But right now I am working on v. Vector v, I need to know that angle of rotation.
Well it is 90… 180… 270, but I did not quite make it to the south line. It is 270
minus 27 degrees. So again standard position rotation of 0, 90, 180, not quite to 270.
I am shy of 270 by 27 degrees. So this angle of theta is going to be 270-27 which is 243
degrees. Let me just double check that. I don’t want to make a mistake. Good! So it
is going to be 50 times cosine of 243 i… plus again the magnitude which is… I don’t
need these magnitude symbols in here any more. 50 times the sine of 243 degrees j. Then because
I can’t do that off the top of my head, that is going to come out to be… Where am I at?
There we go. That is going to be -22.7i plus… or minus:)… 45.55j. My book likes to use
the i j components for horizontal and vertical components. Now we need to find vector w.
So w is going to be… its magnitude which is 80 times the cosine of theta which is 160
degrees i plus again 80 sine 160 j. And then vector w is equal to therefore -75.18i+27.36j.
Now at this point we can stop for a second and just make sure that these numbers seem
to agree with our drawing. Our blue vector has a negative ‘a’ value and a positive ‘b’
value and you can see it is moving to the left and up. So that should be a negative
‘a’ and a positive ‘b’. I also have my yellow vector going to the left and down, so both
my ‘a’ and ‘b’ values should be negative. And indeed they are. So it looks like we are
ok here. The resultant vector is simply now the result of adding these together. The resultant
vector is going to be taking these two and adding them together and w+v… that vector…
the resultant vector is going to be -22.7 plus -75.18, or just minus of course, and
that comes out to be -97.85i… plus a negative…. I don’t usually write that. I don’t know why
I have that in my notes. -17.19j. Again, once you have your horizontal and vertical components
you can simply just add the values of ‘a’… add the horizontal components… add the vertical
components ‘b’ and voila. And does these at least seem to make sense? Well if you draw
a pretty decent diagram, we have a negative ‘a’ value and negative ‘b’ value. That does
appear to be like my drawing where it is slanting down to the left. Again using my reasonably
drawn diagram to verify the reasonableness of my numbers. It is also verifying that with
a negative ‘a’ and a negative ‘b’ that we are in quadrant 3… if the start on the origin.
Ok. Let me get this erased and let’s move on to trying to find the magnitude and direction
of this final vector. But I need more room! So here’s the resulting vector, let’s find
its magnitude. So we’re gonna apply our double absolute value symbol, or whatever this thing
is called, well magnitude of a vector symbol. And that is when we have our vector in component
form we’re just going to work out Pythagorean theorem basically so it’s going to be (-97.85)^2+(-17.19)^2….=99.3. Okay,
what about the final direction? Well, tangent of theta, which is equal to b over a, I’ve
been doing the same kind of conversion from when you went from, let’s say polar coordinates
to rectangular coordinates, and way back when you first started your study in trigonometry,
is going to be b, -17.19, over a, which is -97.85, that is the tangent of theta is equal
to, well, how about we just do this? Theta is equal to the inverse tangent of (-17.19/-97.85).
Of course these negative signs are going to cancel out, and the inverse tangent of that
comes out to be ten degrees. Now that’s all find and dandy, but ten degrees is not in
quadrant three, so again, the drawing are important because they not only let’s you
verify your numbers to the reasononablist as you come along, but now you’re like, okay,
the calculator said its a 10 degree direction, but 10 degrees would be over in quadrant one.
Since we want to be in quadrant three, which means we need to be just past 180 degrees,
and remember inverse tangent will give you a reference angle, if it has a negative, you
can just drop it, I mean, it means something but as far as reference angles, yo can only
have positive reference angles. So actual direction is theta is equal to 180 degrees,
that’s right here, plus another 10. or of course, roughly, rounded off, 190 degrees.
So there is our magnitude and direction of our resultant vector. Let’s go on and do a
couple more examples, BAM! Hey, P.S. 190 degrees, that is south, 80 degrees west. Let’s do a
couple of examples that won’t take so darn long! How bout this one? A two thousand pound
object, probably a car, on a 7.8 degree incline, what amount of force would it take to move
the object up the ramp? Well, let’s see, how bout we have a ramp that has a 7.8 degree
incline, on that ramp, we’ve got something, an object that weighs two thousand pounds.
And of course, you know we need to avoid friction, so it’s just a glassy surface, like ice, no
friction. And that object weighs two thousand pound. Okay well what does gravity do? Gravity
doesn’t pull us down at an angle, gravity pulls us straight down to the earth, so we
have a vector coming down here that represents gravity, and that gravitational force is two
thousand pounds. Okay now, there’s a couple of things going on here. We have the fact
that you want to push this object up the ramp which is going to start at you know break
even, basically how much force does it take to get this thing to be stationary. And that’s
just anything above that is going to push it up the ramp, then you know what kind of
weight is this item actually placing on the ramp? Well you know, when you get on the scale,
your weight is pulled straight down in a perpendicular fashion, so the weigh tof this object is actually
being placed, you know, perpendicular to the ramp. So there’s a vector here, which is directly
how much weight that ramp is feeling from the object itself, and then there are arrows
going in the opposite direction as well, then you have this other line segment that we want
which is right here. Now what i’m setting up is, we have a vertical direction of gravity,
we have the perpendicular angle here, to you know, measure the actual weight of that object
s pressing into the ramp. I mean, if that ramp was really really steep, that object
would just fly down the ramp and would hardly feel any of that weight at all because so
much of it would be going down the side or being cantered off. And by the way, if you
are a physics teacher, or you really dance in physics, please excuse some of my explanations
of this, because I do only do vectors for some of these sections and teach precalculus
so I don’t have a serious in depth knowledge of vectors, just what my textbook covers.
But we are going to… So you’re learning from someone who just admitted that they maybe
don’t now how to read the thing? hmm… Does anybody know everything? Ok. At any rate.
So we are going to make this a right angle and down here a right angle. Look at these
perpendicular components with the gravity vector. Alright. So this orange vector is
going to be the actual weight of the object pressing into the ramp. Then this little segment
is going to be the amount of force and this is what we are interested in. Hmm… Let’s
call this… I don’t know, vector ‘a’. That little bit of a length there is parallel.
That line segment is parallel to the ramp because I have alternate interior angles that
are congruent. I have a perpendicular angle here, I have a perpendicular angle here, I
have two lines being intersected… blah blah blah. So I have a little right triangle in
this corner. Vertical gravity. The force perpendicular to the actual surface of the ramp is the weight
the object is pressing into the ramp… that is what it is supporting. And then we have
some force here at a slant, the same 7.8 degrees, that we have for the angle of the ramp itself.
If we find the length of that vector, we are going to find the force it takes to keep this
object steady. And anything greater than that is going to move the object up the ramp. So
if this is 7.8 degrees here, then this is also going to be right here…. Let’s put
this in white as well. Right here, this little angle is also going to be 7.8 degrees. So
this 7.8… I just kind of also taken this triangle and moved it… rotating it and flipping
it down kind of sort of. So 7.8 here and 7.8 here. Relating to the 7.8 this side that we
are interested in that is parallel to the ramp is the opposite leg. And the 2000 is
acting as the hypotenuse to this right triangle. So what do you have and what do you want?
I want the opposite side and I have the hypotenuse which means we need the sine function. So
the sine of 7.8 degrees is equal to the opposite side of ‘a’ over the hypotenuse of 2000 pounds.
So ‘a’ is going to be equal to 2000 times sine of 7.8. Therefore ‘a’ ends up coming
out to be 271.4 pounds. It is kind of bothering me that I am using ‘a’ which is supposed to
be horizontal component so let me replace it with the letter f for force. 271.4 pounds
and that is how much force we need to keep the object steady. But the question says how
much force does it take to move it up the ramp. Well if that how much force it takes
to keep in balance or equilibrium, which is the concept we are going to talk about down
here, I need a force that is greater than 271.4 pounds to start getting it to move up
the ramp. Any force greater than that will push it up the ramp. Assuming these forces
act…. This is another question by the way. Assuming these forces act upon the same object,
here is force vector oneand force vector 2 is. Now I’m giving you these
component vectors in shorthand. This is the way my Trigonometry book used to have the
vectors in there. My new book is always using the i’s and j’s for horizontal and vertical
components. The pointy brackets are just indicating the difference between plotting a vector and
plotting a point. Well first of all it is a ray, and secondly you don’t have to have
the initial point on the origin. Well, what do we need to put those two vectors into equilibrium.
We need a third vector that when we add all three up we end up with a zero vector. There
is zero force being applied in a horizontal or vertical direction. So force 1 plus force
2 is going to be equal to… How do you add vectors? You just simply add the ‘a’ values
so 2 plus -4, and you add the ‘b’ values -7 plus 3. That comes out to be. Now
what vector could I add toto get a resultant vector of… a Zero Vector.
All I have to do is change the signs. So this vector ofwill put these forces into
equilibrium. Alright. Mr. Tarrou and I have one more example. Now for this last example,
a plane is traveling 245 miles per hour… that is air speed… at a bearing of North
63 degrees West. Now the reason I say this is air speed is because we are going to have
a second force happening which is wind. Now if you are flying into the wind your air speed
is going to be faster than your ground speed. Because you are flying against the wind and
in relation to that you are going a certain speed. But that wind is actually slowing you
down. Or maybe it is a tail wind and it is speeding you up. The wind is blowing at 10
miles per hour in the direction South 10 degrees East. (matching work) Approximate the ground
speed and we are going to also be approximating the direction. This may sound very familiar
to the second example that I just did and indeed it is. So I am not going to actually
walk you through this problem. I am just going to give you the solutions, but give you some
time first to try them on your own. So copy that down and let’s see what the diagram looks
like. I hope this is what your diagram looks like. Now I am drawing my vectors End to End.
You can draw your plane vector and your wind vector leaving the same point and draw it
in a parallelogram fashion as trying to visualize what the resultant vector looks like. I drew
mine End to End. So we have a plane vector with an amplitude of 245mph. We have a wind
vector a force or magnitude, or length, of 10 mph. And with magnitude we need direction.
Now not bearing direction, but Standard Position rotation. So if it is North 63 degrees West,
that is actually going to be an angle measure for the plane vector of 90 plus another 63
degrees which is going to be 153 degrees. The wind is going to have…. Now see where
the wind is? That wind angle has to rotate all around. And from zero you go to 90…
180… 270… and an additional 10 degrees. So the direction for the wind is 280 degrees.
We have our diagram. We need to change our vectors from magnitude and direction into
the horizontal and vertical components. Let’s see if you can do that. One for the plane
and of course one for the wind. So to go from our magnitude and direction into our horizontal
and vertical components again it is the magnitude of the vector times the cosine of its angle
times vector ‘i’… the horizontal component… plus 245 which is the magnitude of the vector….
Oops… The sine of the angle. We get the horizontal and vertical components of
I am using my shorthand notation. Not mine, but my old textbook’s notation because I am
getting tired of writing ‘i’ and ‘j’ all the time. Now look at the white vector. It is
going to the left and up. So we should have been expecting a negative ‘a’ value and a
positive ‘b’ value. And we have it. The wind vector going down and to the right a little
bit, we should be expecting a positive ‘a’ and a negative ‘b’. And indeed we have that.
That is equal to the magnitude of the vector times the cosine of its direction ‘i’ plus
the magnitude of the vector.. in this case it is wind speed of course… sine of the
angle direction times vector j, the vertical component. And you get the short hand version
ofWell now that you have the horizontal and vertical components of each
vector it should be pretty straight forward to find the resultant. We are just going to
add those values of ‘a’ and ‘b’ together. And then for theta, again one more time, this
is drawn and you can see that it looks it is in quadrant 2. Hopefully the resultant
has negative ‘a’ and positive ‘b’ to verify that it is quadrant 2. So when you use the
tangent function to find out what the angle measure, make sure that you don’t take that
number out of the calculator. You need to convert it into something that reflects the
fact that your vector is in quadrant 2 and not in quadrant… umm… What other quadrant
is Tangent negative in? So… The resultant of two vectors when you have them in component
form like we have here again is add the ‘a’, add the ‘b’. So our resultant vector is
The magnitude of that vector, again if you had your vector in component form, you are
just working through the Pythagorean Theorem. So our final speed… our ground speed is
239.1 miles per hour. That should make sense because you see the way this vector is drawn
with the plane. The wind is fighting it a little bit which is going to slow that plane
down some. As far as the final direction of this resultant vector figuring out what theta
is going to be equal to, the tangent of theta is b over a… or 101.35/-216.56. After you
do the inverse tangent function you get negative 25.1 degrees. That is because this vector
is drawn in quadrant 2. In quadrant 2 we have negative ‘a’ and positive ‘b’, or way back
in Trig it would be negative x’s and positive y’s. When you set up that ratio of y/x or
now in vectors b/a, if those signs are different you get a negative answer of course or a negative
ratio. That happens as well in quadrant 4. So while this is a good number, it is not
going to be our final answer. Negative 25.1 degrees. Lose the negative and you have a
reference angle. The actual answer is going to be… I want to rotate…. Remember reference
angles are an acute measurement off the x axis if you will if you want to place it there
for convenience. I want to rotate 180 degrees and then back off the reference angle because
I am in quadrant 2, of 25.1 degrees. So 180-25.1 is your angle of rotation of 154.9 degrees.
Now if the whole problem starts off in bearings your teacher is probably going to want that
final answer in bearings as well. So with bearings you either reference North or South
first. And clearly quadrant 2 we are closer to the direction of North. So 154.9 is North
64.9 degrees West, or take that… Let’s see here. We have 90 and I do I want to work this
out? Oh! This reference angle is 25.1 degrees, so I can do 90, which is the difference between
North and West… take 90 and subtract away the 25.1 and you get your 64.9. So WHOO!!
That was a lot of examples. I hope they help. I am Mr. Tarrou. BAM! Now go do your homework:)